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Evaluation of double integrals

We know that the double integral over the region 'R' of a function f(x, y) is

$$\int\!\!\!\int\limits_R {f(x,y)\;} dx\;dy \;\;\;\;\;\;\;\;\;\;\;.........{\rm{(1)}}$$

Case (i)
Now let 'R' be the region bounded by the lines x = c1, x = c2, y = c3 and y = c4 where c1, c2, c3, c4 are constants.

Since we know the region of integration 'R' the double integral $$\int\!\!\!\int\limits_R {f(x,y)\;} dx\;dy$$ can be written as

$$\int\limits_{c3}^{c4} {\int\limits_{c1}^{c2} {f(x,y)\;} } dx\;dy \;\;\;\;\;\;\;\;\;\;\;.........{\rm{(2)}}$$

(Here we replace ‘R’ by putting the limits for both x and y)

Note

For 'n' subregions, we have 'n' points P1, P2, ……. Pn. Let f (P1), f (P2), ……. f (Pn) be the values of the functions at the points Pi. Now from the sum of the products f(Pi) si.

$$\eqalign{ \int\limits_{c3}^{c4} {\int\limits_{c1}^{c2} {f(x,y)\;dx\;dy} } & = \int\limits_{c3}^{c4} {\int\limits_{c1}^{c2} {f(x,y)\;dx\;dy} } \cr \int\limits_{c3}^{c4} {\int\limits_{c1}^{c2} {f(x,y)\;dx\;dy} } & = \int\limits_{c1}^{c2} {\int\limits_{c3}^{c4} {f(x,y)\;dx\;dy} } \cr} $$

In L.H.S., the order of integration is first with respect to x and then with respect to y.
In R.H.S. the order of integration is first with respect to y and then with respect to x.
Case (ii)

Consider the double integral $$\int\limits_{c1}^{c2} {\int\limits_{x1}^{x2} {f(x,y)\;} } dx\;dy$$

Suppose x1 and x2 are functions of y say x1 = f(y), x2 = (y) and c1 and c2 are constants then the region of integration is bounded by the curves x1 = f(y) and x2 =  (y) and the lines y = c1 and y = c2.

Fig 1

This region is shown in figure. Here we integrate f(x, y) first w.r.t. x. Keeping pression w.r.t.y. i.e., first integration is along the horizontal strip PQ and then slide this strip PQ vertically.
Case (iii)

Now consider the double integral $$\int\limits_{c1}^{c2} {\int\limits_{y1}^{y2} {f(x,y)\;} } dx\;dy$$

Suppose y1 and y2 are functions of x say y1 = f(x), y2 =  (x) and the lines x = c1 and x = c2.

Fig 2

This region is shown in figure. Here we integrate f(x, y) first w.r.t. y keeping x as a constant and integrate the resulting expressions w.r.t. x. In other words first integration is along the vertical strip RS and then slide this strip RS horizontally.

Consider the double integral $$\int\limits_{c1}^{c2} {\int\limits_{\phi 1(y)}^{\phi 2(y)} {f(x,y)\;dx\;dy} } $$

Note

Here we note that when the limits of the inner integral remain the function of y, our first integration should be w.r.t. x.

Example 1 :

Evaluation of definite integral $$\int\limits_0^{{\pi \over 2}} {{{\sin }^m}} x.{\cos ^n}x\;dx$$ when both m and n are positive integers

Solution

Introduce the notation

$$\mathop I\nolimits_{m,n} = \int\limits_0^{{\pi \over 2}} {{{\sin }^m}x.{{\cos }^n}x} \;dx \;\;\;\;\;\;\;\;\;\;\;.........{\rm{(1)}}$$

And using the reduction formula, we obtain

$$\eqalign{ \mathop I\nolimits_{m,n} & = \int\limits_0^{{\pi \over 2}} {{{\sin }^m}x.{{\cos }^n}x} \;dx \cr & = \left[ { - {{{{\sin }^{m - 1}}x.{{\cos }^{n + 1}}} \over {m + n}}\mathop \;\limits_0^{{\pi \over 2}} + {{m - 1} \over {m + n}}} \right]\int\limits_0^{{\pi \over 2}} {{{\sin }^{m - 2}}x.{{\cos }^n}x\;dx} \cr} $$

Since the first term on R.H.S is zero at both the limits, we get

$$\mathop I\nolimits_{m,n} = {{m - 1} \over {m + 1}}.\mathop I\nolimits_{m - 2,n} \;\;\;\;\;\;\;\;\;\;\;.........{\rm{(2)}}$$

Using this recurrence relation, we obtain

$$\eqalign{ \mathop I\nolimits_{m - 2,n} & = {{m - 5} \over {m + n - 2}}.\mathop I\nolimits_{m - 4,n} \;\;\;\;\;\;\;\;\;\;\;.........{\rm{(3)}} \cr \mathop I\nolimits_{m - 4,n} & = {{m - 5} \over {m + n - 4}}.\mathop I\nolimits_{m - 6,n} \cr - - - - & - - - - - - - - - - - - - - - \cr \mathop I\nolimits_{3,n} & = {2 \over {3 + n}}.\mathop I\nolimits_{1,n} \;When\;m\;is\;odd \cr \mathop I\nolimits_{2,n} & = {1 \over {2 + n}}.\mathop I\nolimits_{0,n} \;When\;m\;is\;even \cr}$$

Observe that

$$\eqalign{ \mathop I\nolimits_{1,n} & = \int\limits_0^{{\pi \over 2}} {\sin x.{{\cos }^n}x\;dx} = - {{{{\cos }^{n + 1}}x} \over {n + 1}}_0^{{\pi \over 2}} + 0 = {1 \over {n = 1}} \;\;\;\;\;\;\;\;\;\;\;.........{\rm{(4)}} \cr & and \cr \mathop I\nolimits_{0,n} & = \int\limits_0^{{\pi \over 2}} {{{\cos }^n}x\;dx} \cr}$$

Substuting these values, we have (a) When m is odd and n may be odd or even

$$\eqalign{ {I_{m,n}} & = {{(m - 1)} \over {(m + n)}}.{{(m - 3)} \over {(m + n - 2)}} \times {{(m - 5)} \over {(m + n - 4)}}......{2 \over {3 + n}}.{1 \over {1 + n}} \cr}$$

rewritting

$$\eqalign{{I_{m,n}} & = {{(m - 1)(m - 3)(m - 5)......} \over {(m + n)(m + n - 2){\rm{ }}......}} \times {1 \over {2 + n}}\left[ {{{n - 1} \over n}{\rm{ }}{\rm{. }}{{n - 3} \over {n - 2}}{\rm{ }}........{\rm{ }}{2 \over 3}} \right] \cr} $$

(b) When m is even and n is odd

$$\eqalign{{I_{m,n}} & = {{(m - 1)(m - 3)(m - 5)......{\rm{ 2 }}{\rm{. 1}}} \over {(m + n)(m + n - 2)(m + n - 4){\rm{ }}......{\rm{ (}}3 + n{\rm{) (}}1 + n{\rm{)}}}}. \times \left[ {{{(n - 1)(n - 3)(n - 5)......} \over {(n - 1)(n - 3)(n - 5)......}}} \right] \cr} $$

$$\eqalign{{I_{m,n}} = {{(m - 1)(m - 3){\rm{ }}......{\rm{ 1}}} \over {(m + n)(m + n - 2){\rm{ }}......{\rm{ (n + 2)}}}} \times {1 \over {2 + n}}\left[ {{{n - 1} \over n}{\rm{ }}{\rm{. }}{{n - 3} \over {n - 2}}{\rm{ }}........{\rm{ }}{2 \over 3}} \right] \cr} $$